1. **Problem 1: Finding the lowest mark for Grade A** Ben needs an average of at least 70 over 5 tests. Formula for average: $$\text{Average} = \frac{\text{Sum of all marks}}{\text{Number of tests}}$$ Given average for first 4 tests is 68, so sum of first 4 tests is $$4 \times 68 = 272$$. Let the mark for the 5th test be $x$. We want: $$\frac{272 + x}{5} \geq 70$$ Multiply both sides by 5: $$272 + x \geq 350$$ Subtract 272 from both sides: $$x \geq 350 - 272$$ $$x \geq 78$$ So, the lowest mark Ben can get on the fifth test to obtain Grade A is **78**. 2. **Problem 2: Standard deviation of masses from two machines** Standard deviation formula for a sample: $$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2}$$ where $n$ is the number of samples, $x_i$ are the data points, and $\bar{x}$ is the sample mean. **Machine A:** Data: 196, 198, 198, 199, 200, 200, 201, 201, 202, 205 Calculate mean: $$\bar{x}_A = \frac{196 + 198 + 198 + 199 + 200 + 200 + 201 + 201 + 202 + 205}{10} = \frac{1990}{10} = 199$$ Calculate squared deviations and sum: $$(196-199)^2 = 9$$ $$(198-199)^2 = 1$$ (twice) total 2 $$(199-199)^2 = 0$$ $$(200-199)^2 = 1$$ (twice) total 2 $$(201-199)^2 = 4$$ (twice) total 8 $$(202-199)^2 = 9$$ $$(205-199)^2 = 36$$ Sum of squared deviations: $$9 + 2 + 0 + 2 + 8 + 9 + 36 = 66$$ Calculate standard deviation: $$s_A = \sqrt{\frac{66}{10-1}} = \sqrt{\frac{66}{9}} = \sqrt{7.33} \approx 2.71$$ **Machine B:** Data: 192, 194, 195, 198, 200, 201, 203, 204, 206, 207 Calculate mean: $$\bar{x}_B = \frac{192 + 194 + 195 + 198 + 200 + 201 + 203 + 204 + 206 + 207}{10} = \frac{2000}{10} = 200$$ Calculate squared deviations and sum: $$(192-200)^2 = 64$$ $$(194-200)^2 = 36$$ $$(195-200)^2 = 25$$ $$(198-200)^2 = 4$$ $$(200-200)^2 = 0$$ $$(201-200)^2 = 1$$ $$(203-200)^2 = 9$$ $$(204-200)^2 = 16$$ $$(206-200)^2 = 36$$ $$(207-200)^2 = 49$$ Sum of squared deviations: $$64 + 36 + 25 + 4 + 0 + 1 + 9 + 16 + 36 + 49 = 240$$ Calculate standard deviation: $$s_B = \sqrt{\frac{240}{9}} = \sqrt{26.67} \approx 5.16$$ **Comment:** Machine B has a higher standard deviation, meaning its packet masses vary more than Machine A's. 3. **Problem 3: Median, quartiles, and interquartile range** Data: 52, 61, 78, 49, 47, 79, 54, 58, 62, 73, 72 Step 1: Sort the data: $$47, 49, 52, 54, 58, 61, 62, 72, 73, 78, 79$$ (a) Median (middle value for 11 data points): 6th value $$\text{Median} = 61$$ (b) Lower quartile (median of lower half excluding median): Lower half: $$47, 49, 52, 54, 58$$ Median of lower half (3rd value): $$\text{Lower quartile} = 52$$ (c) Upper quartile (median of upper half excluding median): Upper half: $$62, 72, 73, 78, 79$$ Median of upper half (3rd value): $$\text{Upper quartile} = 73$$ (d) Interquartile range (IQR): $$\text{IQR} = \text{Upper quartile} - \text{Lower quartile} = 73 - 52 = 21$$